On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k Ω. How much was the resistance on the left slot before interchanging the resistances? (a) 910 Ω (b) 990 Ω (c) 505 Ω (d) 550 Ω

Answer: (d). Before: (R₁)/(R₂) = (l)/(100-l). After interchange, balance at l-10: (R₂)/(R₁) = (l-10)/(110-l). So (l)/(100-l) = (110-l)/(l-10) ⇒ l² - 10l = 11000 - 210l + l² ⇒ l = 55 cm. R₁ = 1000 × 55/100 = 550 Ω. So the answer is (d) 550 Ω.

On interchanging the resistances, the balance point of a meter bridge… | JEE Main 2018

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Wheatstone Bridge & Metre Bridge

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k $\Omega$ . How much was the resistance on the left slot before interchanging the resistances?

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Answer: (d)

Before: $\dfrac{R_1}{R_2} = \dfrac{l}{100-l}$ . After interchange, balance at $l-10$ : $\dfrac{R_2}{R_1} = \dfrac{l-10}{110-l}$ .
So $\dfrac{l}{100-l} = \dfrac{110-l}{l-10} \Rightarrow l^2 - 10l = 11000 - 210l + l^2 \Rightarrow l = 55\text{ cm}$ .
$R_1 = 1000 \times \dfrac{55}{100} = 550\ \Omega$ .
So the answer is (d) $550\ \Omega$ .

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